$\{1,\sqrt [3]{7},{\sqrt [3]{7 }}^{2}\}$ is a basis for $Q[\sqrt {2},\sqrt {3},\sqrt [3]{7}]$ over $Q[\sqrt {2},\sqrt{3}]$ I am trying to prove that 
$\{ 1,\sqrt [ 3 ]{ 7 } ,{ \sqrt [ 3 ]{ 7 }  }^{ 2 }\} \quad is\quad a\quad basis\quad for\quad Q[\sqrt { 2 } ,\sqrt { 3 } ,\sqrt [ 3 ]{ 7 } ]\quad over\quad Q[\sqrt { 2 } ,\sqrt { 3 } ]$.
I have proved that the dimension is 3, so I have tried to prove that these elements are linearly independent over the field, but the calculations got too complicated.
How can I do that?
 A: If the dimension is 3 then the degree of the minimal polynomial for $\sqrt[3]{7}$ over $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is 3. Suppose for contradiction that $1,\sqrt[3]{7},$ and $\sqrt[3]{7}^2$ are $\mathbb{Q}(\sqrt{2},\sqrt{3})$-linearly dependant. Then there exist $c_i \in \mathbb{Q}(\sqrt{2},\sqrt{3})$ with at least one $c_i$ non-zero such that $c_1 \sqrt[3]{7}^2 + c_2 \sqrt[3]{7} + c_3 = 0$. However this means $\sqrt[3]{7}$ is a root of  $c_1 x^2 + c_2 x + c_3 \in \mathbb{Q}(\sqrt{2},\sqrt{3})[x] ,$ contradicting the minimality of the minimal polynomial.

This process works in general. If $\mathbb{F}$ is a field and $\alpha$ is algebraic over $\mathbb{F}$ then $|\mathbb{F}(\alpha):\mathbb{F}| = n$ where $n$ is the degree of the minimal polynomial. You can then show that $\{1,\alpha, \ldots , \alpha^{n-1}\}$ is a basis for $\mathbb{F}(\alpha)$ over $\mathbb{F}$ by showing they are linearly independent using a similar argument to above. This is actually how you prove that $|\mathbb{F}(\alpha):\mathbb{F}|$  is the degree of the minimal polynomial.
